Show how am MDP with a reward function R(s, a, s’) can be transformed into a different MDP with reward function R(s, a), such that optimal policies in the new MDP correspond exactly to optimal policies in the original MDP

U(s) = maxa[R0

(s, a) + γ

1

2

P

pre T

0

(s, a, pre)(maxb[R0

(pre, b) + γ

1

2

P

s

0 T

0

(pre, b, s0

) ∗ U(s

0

))]]

U(s) = maxa[

P

s

0 T(s, a, s0

)(R(s, a, s0

) + γU(s

0

)]

U(s) = R0

(s) + γ

1

2 maxa[

P

post T

0

(s, a, post)(R0

(post) + γ

1

2 maxb[

P

s

0 T

0

(post, b, s0

)U(s

0

))]]

U(s) = maxa[R(s, a) + γ

P

s

0 T(s, a, s0

)U(s

0

)]

Explanation:

MDPs

MDPs can formulated with a reward function R(s), R(s, a) that depends on the action taken or R(s, a, s’) that

depends on the action taken and outcome state.

To Show how am MDP with a reward function R(s, a, s’) can be transformed into a different MDP with reward

function R(s, a), such that optimal policies in the new MDP correspond exactly to optimal policies in the

original MDP.

One solution is to define a ’pre-state’ pre(s, a, s’) for every s, a, s’ such that executing a in s leads not to s’

but to pre(s, a, s’). From the pre-state there is only one action b that always leads to s’. Let the new MDP

have transition T’, reward R’, and discount γ

0

.

T

0

(s, a, pre(s, a, s0

)) = T(s, a, s0

)

T

0

(pre(s, a, s0

), b, s0

) = 1

R0

(s, a) = 0

R0

(pre(s, a, s0

), b) = γ

− 1

2 R(s, a, s0

)

γ

0 = γ

1

2

Then, using pre as shorthand for pre(s, a, s’):

U(s) = maxa[R0

(s, a) + γ

1

2

P

pre T

0

(s, a, pre)(maxb[R0

(pre, b) + γ

1

2

P

s

0 T

0

(pre, b, s0

) ∗ U(s

0

))]]

U(s) = maxa[

P

s

0 T(s, a, s0

)(R(s, a, s0

) + γU(s

0

)]

Now do the same to convert MDPs with R(s, a) into MDPs with R(s).

Similar to part (c), create a state post(s, a) for every s, a such that

T

0

(s, a, post(s, a, s0

)) = 1

T

0

(post(s, a, s0

), b, s0

) = T(s, a, s0

)

R0

(s) = 0

R0

(post(s, a, s0

)) = γ

− 1

2 R(s, a)

γ

0 = γ

1

2

Then, using post as shorthand for post(s, a, s’):

U(s) = R0

(s) + γ

1

2 maxa[

P

post T

0

(s, a, post)(R0

(post) + γ

1

2 maxb[

P

s

0 T

0

(post, b, s0

)U(s

0

))]]

U(s) = maxa[R(s, a) + γ

P

s

0 T(s, a, s0

)U(s

0

)]

3

hi

explanation:

answer:

an example of solubility is salt or sugar dissolving in water.

rate as brainliest plz